LeetCode 63: Unique Paths II

Problem Description

LeetCode Problem 63

You are given an m x n integer array grid. There is a robot initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time.

An obstacle and space are marked as 1 or 0 respectively in grid. A path that the robot takes cannot include any square that is an obstacle.

Return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The testcases are generated so that the answer will be less than or equal to 2 * 109.

Example 1:

Input: obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
Output: 2
Explanation: There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:

Right -> Right -> Down -> Down
Down -> Down -> Right -> Right

Example 2:

Input: obstacleGrid = [[0,1],[0,0]]
Output: 1

Constraints:

• m == obstacleGrid.length
• n == obstacleGrid[i].length
• 1 <= m, n <= 100
• obstacleGrid[i][j] is 0 or 1.

Difficulty: Medium

Tags: array, dynamic programming, matrix

Rating: 94.50%

Solution

Here’s my Python solution to this problem:

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        m, n = len(obstacleGrid), len(obstacleGrid[0])
        dp = [[0] * n for i in range(m)]

        for i in range(m):
            if obstacleGrid[i][0] == 1:
                break
            else:
                dp[i][0] = 1

        for j in range(n):
            if obstacleGrid[0][j] == 1:
                break
            else:
                dp[0][j] = 1

        for i in range(1, m):
            for j in range(1, n):
                if obstacleGrid[i][j] == 1:
                    continue
                dp[i][j] = dp[i-1][j] + dp[i][j-1]
        
        return dp[m-1][n-1]

Complexity Analysis

The solution has the following complexity characteristics:

• Time Complexity: $O(n²)$
• Space Complexity: $O(n)$

Note: This is an automated analysis and may not capture all edge cases or specific algorithmic optimizations.